z(1+20/7)=540

Solution for z(1+20/7)=540 equation:

z(1+20/7)=540

We move all terms to the left:

z(1+20/7)-(540)=0

We add all the numbers together, and all the variables

z(20/7+1)-540=0

We multiply parentheses

20z^2+z-540=0

a = 20; b = 1; c = -540;
Δ = b2-4ac
Δ = 12-4·20·(-540)
Δ = 43201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{43201}}{2*20}=\frac{-1-\sqrt{43201}}{40}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{43201}}{2*20}=\frac{-1+\sqrt{43201}}{40}
$